Unique Expansions of Real Numbers

Transcription

1 ESI The Erwin Schrödinger International Boltzmanngasse 9 Institute for Mathematical Physics A-1090 Wien, Austria Unique Expansions of Real Numbers Martijn de Vries Vilmos Komornik Vienna, Preprint ESI 1810 (2006) June 29, 2006 Supported by the Austrian Federal Ministry of Education, Science and Culture Available via

2 UNIQUE EXPANSIONS OF REAL NUMBERS MARTIJN DE VRIES AND VILMOS KOMORNIK Abstract. It was discovered some years ago that there exist non-integer real numbers q > 1 for which only one sequence (c i ) of integers 0 c i < q satisfies the equality P c iq i = 1. The set U of such univoque numbers has a rich topological structure, and its study revealed a number of unexpected connections with measure theory, fractals, ergodic theory and Diophantine approximation. For each fixed q > 1 consider the set U q of real numbers x having a unique expansion of the form P c iq i = x with integers 0 c i < q. We carry out a detailed topological study of these sets. In particular, we characterize their closures, and we determine those bases q for which U q is closed or even a Cantor set. 1. Introduction Following a seminal paper of Rényi [R] many works were devoted to probabilistic, measure-theoretical and number theoretical aspects of developments in non-integer bases; see, e.g., Frougny and Solomyak [FS], Pethő and Tichy [PT], Schmidt [Sc]. A new research field was opened when Erdős, Horváth and Joó [EHJ] discovered many non-integer real numbers q > 1 for which only one sequence (c i ) of integers 0 c i < q satisfies the equality c i q = 1. i (They considered only the case 1 < q < 2.) Subsequently, the set U of such univoque numbers was characterized in [EJK1], [KL3], its smallest element was determined in [KL1], and its topological structure was described in [KL3]. On the other hand, the investigation of numbers q for which there exist continuum many such sequences, including sequences containing all possible finite variations of the integers 0 c < q revealed close connections to Diophantine approximations; see, e.g., [EJK1], [EJK3], [EK], [KLP], Borwein and Hare [BH1], [BH2], Komatsu [K], and Sidorov [Si2]. For any fixed real number q > 1, we may also introduce the set U q of real numbers x having exactly one expansion of the form c i q = x i where the integer coefficients c i are subject to the conditions 0 c i < q. If q is an integer, these sets are well known. However, their structure is more complex if q is a non-integer, see, e.g., Daróczy and Kátai [DK1], [DK2], Glendinning and Sidorov [GS], and Kallós [K1], [K2]. The purpose of this paper is to give a complete topological description of the sets U q : they have a different nature for different Date: June 29, Mathematics Subject Classification. Primary:11A63, Secondary:11B83. Key words and phrases. Greedy expansions, beta-expansions, univoque sequences, univoque sets, Cantor sets. 1

3 2 MARTIJN DE VRIES AND VILMOS KOMORNIK classes of the numbers q. Our investigations also provide new results concerning the univoque set U. In order to state our results we need to introduce some notation and terminology. In this paper a sequence always means a sequence of nonnegative integers. A sequence is called infinite if it contains infinitely many nonzero elements; otherwise it is called finite. Given a real number q > 1, an expansion in base q of a number x is a sequence (c i ) such that 0 c i < q for all i 1 and x = This definition only makes sense if x belongs to the interval [ J := 0, q 1 ] q 1 where q denotes the upper integer part of q. Note that [0, 1] J for all q > 1. A sequence (c i ) satisfying 0 c i < q for each i 1 is called univoque in base q if c i x = q i is an element of U q. The greedy expansion (b i (x)) = (b i ) of a number x J in base q is the largest expansion of x in lexicographical order. It is well known that the greedy expansion of any x J exists; [P], [EJK1], [EJK2]. A sequence (b i ) is called greedy in base q if (b i ) is the greedy expansion of x = The quasi-greedy expansion (a i (x)) = (a i ) of a number x J \ {0} in base q is the largest infinite expansion of x in lexicographical order. Observe that we have to exclude the number 0 since there do not exist infinite expansions of x = 0 at all. On the other hand, the largest infinite expansion of any x J \ {0} exists, as we shall prove in the next section. In order to simplify some statements below, the quasi-greedy expansion of the number 0 J is defined to be 0 = Note that this is the only expansion of x = 0. A sequence (a i ) is called quasi-greedy in base q if (a i ) is the quasi-greedy expansion of x = We denote the quasi-greedy expansion of the number 1 in base q by (α i ). Since α 1 = q 1 the digits of an expansion (c i ) satisfy b i q i. a i q i. c i {0,..., α 1 } for all i 1. Hence, we consider expansions with coefficients or digits in the alphabet A := {0,..., α 1 } of numbers x [0, α 1 /(q 1)]. Of course, whether a sequence is univoque, greedy or quasi-greedy depends on the base q. However, if q is understood, we simply speak of univoque sequences and (quasi)-greedy sequences. Furthermore, we shall write a := α 1 a (a A), unless stated otherwise. Finally, we set a + := a + 1 and a := a 1, where a A. The following important theorem, which is essentially due to Parry (see [P]), plays a crucial role in the proofs of our main results: Theorem 1.1. Fix q > 1. c i q i.

4 UNIQUE EXPANSIONS OF REAL NUMBERS 3 (i) A sequence (b i ) = b 1 b 2... {0,..., α 1 } N is a greedy sequence in base q if and only if b n+1 b n+2... < α 1 α 2... whenever b n < α 1. (ii) A sequence (c i ) = c 1 c 2... {0,..., α 1 } N is a univoque sequence in base q if and only if and c n+1 c n+2... < α 1 α 2... whenever c n < α 1 c n+1 c n+2... < α 1 α 2... whenever c n > 0. Note that if q U, then (α i ) is the unique expansion of 1 in base q. Hence, replacing the sequence (c i ) in Theorem 1.1 (ii) by the sequence (α i ), one obtains a lexicographical characterization of U. Recently, the authors of [KL3] studied the topological structure of the set U. In particular, they showed that U is not closed and they obtained the following characterization of its closure U: Theorem 1.2. q U if and only if the quasi-greedy expansion of the number 1 in base q satisfies α k+1 α k+2... < α 1 α 2... for all k 1. Remark. In the definition of U given in [KL3] the integers were excluded; however, U is the same in both cases. Our definition simplifies some statements. For example it will follow from the theorems below that where U q denotes the closure of U q. U q = U q q (1, ) \ U Now we are ready to state our main results. Theorem 1.3. Let q U and x J. Denote the quasi-greedy expansion of x by (a i ). Then, x U q a n+1 a n+2... α 1 α 2... whenever a n > 0. Theorem 1.4. Suppose that q U. Then, (i) U q \ U q = ℵ 0 and U q \ U q is dense in U q. (ii) If q U, then each element x U q \ U q has 2 expansions. (iii) If q U \ U, then each element x U q \ U q has ℵ 0 expansions. Remarks. Our proof of part (i) yields the following more precise results where for q U we set A q = { x U q \ U q : x has a finite greedy expansion } and B q = { x U q \ U q : x has an infinite greedy expansion } : If q U \ N, then both A q and B q are countably infinite and dense in U q. Moreover, the greedy expansion of a number x B q ends with α 1 α If q = 2, 3,..., then B q =. For each x U q \ U q, the proof of parts (ii) and (iii) also provide the list of all expansions of x in terms of its greedy expansion.

5 4 MARTIJN DE VRIES AND VILMOS KOMORNIK Motivated by Theorem 1.3, we introduce for a general real number q > 1, the set V q, defined by { } V q = x J : a n+1 (x)a n+2 (x)... α 1 α 2... whenever a n (x) > 0. It follows from the above theorems that U q U q = V q if q U. It is natural to study the relationship between the sets U q, U q and V q in case q / U. In order to do so, we introduce the set V, consisting of those numbers q > 1, for which the quasi-greedy expansion of the number 1 in base q satisfies α k+1 α k+2... α 1 α 2... for each k 1. It follows from Theorem 1.1 and Theorem 1.2 that U U V and U q V q for all q > 1. The following results imply that U q is closed if q / U and that the set V q is closed for each number q > 1. Theorem 1.5. Suppose that q V \ U. Then, (i) The sets U q and V q are closed. (ii) V q \ U q = ℵ 0 and V q \ U q is a discrete set, dense in V q. (iii) Each element x V q \ U q has ℵ 0 expansions, and a finite greedy expansion. Remark. Our proof also provides the list of all expansions of all elements x V q \U q. Theorem 1.6. Suppose that q (1, ) \ V. Then, U q = U q = V q. Remarks. In view of the above results, Theorem 1.1 already gives us a lexicographical characterization of U q in case q / U because in this case U q = U q. It is well-known that the set U has Lebesgue measure zero; [EHJ], [KK]. In [KL3] it was shown that the set U \ U is countably infinite. It follows from the above results that U q is closed for almost every q > 1. Let q > 1 be a non-integer. In [DDV] it has been proved that almost every x J has a continuum of expansions in base q (see also [Si1]). It follows from the above results that the set U q has Lebesgue measure zero. Hence, the set U q is nowhere dense for any non-integer q > 1. Let q > 1 be an integer. In this case, the quasi-greedy expansion of 1 in base q is given by (α i ) = α 1 = (q 1). It follows from Theorem 1.1 that J \ U q is countable and each element in J \ U q has only two expansions, one of them being finite while the other one ends with an infinite string of (q 1) s. In [KL1] it was shown that the smallest element of U is given by q 1.787, and the unique expansion of 1 in base q is given by the truncated Thue- Morse sequence (τ i ) = τ 1 τ 2..., which can be defined recursively by setting τ 2 N = 1 for N = 0, 1, 2,... and τ 2 N +i = 1 τ i for 1 i < 2 N, N = 1, 2,.... Subsequently, Glendinning and Sidorov [GS] proved that U q is countable if 1 < q < q and has the cardinality of the continuum if q [q, 2). Moreover, they showed that U q is a set of positive Hausdorff dimension if q < q < 2, and they described a method to compute its Hausdorff dimension (see also [DK2], [K1], [K2]). In the following theorem we characterize those q > 1 for which U q or U q is a Cantor set, i.e., a non-empty closed set having no interior or isolated points. We recall from [KL3] that

6 UNIQUE EXPANSIONS OF REAL NUMBERS 5 V is closed and U is closed from above, U \ U = ℵ 0 and U \ U is dense in U, V \ U = ℵ 0 and V \ U is a discrete set, dense in V. Since the set (1, )\V is open, we can write (1, )\V as the union of countably many disjoint open intervals (q 1, q 2 ): its connected components. Let us denote by L and R the set of left (respectively right) endpoints of the intervals (q 1, q 2 ). Theorem 1.7. (i) L = N (V \ U) and R = V \ U. Hence, R L and (1, ) \ U = (q 1, q 2 ] where the union runs over the connected components (q 1, q 2 ) of (1, ) \ V. (ii) If q {2, 3,...}, then neither U q nor U q is a Cantor set. (iii) If q U \ N, then U q is not a Cantor set, but its closure U q is a Cantor set. (iv) If q (q 1, q 2 ], where (q 1, q 2 ) is a connected component of (1, ) \ V, then U q is a Cantor set if and only if q 1 {3, 4,...} (U \ U). Remark. We also describe the set of endpoints of the connected components (p 1, p 2 ) of (1, ) \ U: denoting by L and R the set of left (respectively right) endpoints of the intervals (p 1, p 2 ), we have see the remarks at the end of Section 7. L = N (U \ U) and R U; The above theorem enables us to give a new characterization of the stable bases, introduced and investigated by Daróczy and Kátai ([DK1], [DK2]). Let us denote by U q and V q the sets of quasi-greedy expansions of the numbers x U q and x V q. We recall that a number q > 1 is stable from above (respectively stable from below) if there exists a number s > q (respectively 1 < s < q) such that U q = U s. Furthermore, we say that an interval I (1, ) is a stability interval if U q = U s for all q, s I. Theorem 1.8. The maximal stability intervals are given by the singletons {q} where q U and the intervals (q 1, q 2 ] where (q 1, q 2 ) is a connected component of (1, ) \ V. Moreover, if q 1 V \ U, then U q = V q 1 for all q (q 1, q 2 ]. Remark. The proof of Theorem 1.8 yields a new characterization of the sets U and V (see Lemma 7.9). We recall from [LM] that a set A {0,..., α 1 } N is called a subshift of finite type if there exists a finite set F k=1 {0,..., α 1} k such that a sequence (c i ) {0,..., α 1 } N belongs to A if and only if each word in F does not appear in (c i ). The following theorem characterizes those q > 1 for which U q is a subshift of finite type. Theorem 1.9. Let q > 1 be a real number. Then, U q is a subshift of finite type q (1, ) \ U. Finally, we determine the cardinality of U q for all q > 1. We recall that for q (1, 2) this has already been done by Glendinning and Sidorov ([GS]), using a different method. Denote by q the smallest element of U (2, 3). The unique expansion of 1 in base q is given in [KL2].

7 6 MARTIJN DE VRIES AND VILMOS KOMORNIK Theorem Let q > 1 be a real number. (i) If q (1, (1 + 5)/2], then U q consists merely of the endpoints of J. (ii) If q ((1 + 5)/2, q ) (2, q ), then U q = ℵ 0. (iii) If q [q, 2] [q, ), then U q = 2 ℵ0. Remark. We also determine the unique expansion of 1 in base q (n) for n {3, 4,...} where q (n) denotes the smallest element of U (n, n + 1) (see the remarks at the end of Section 7). For the reader s convenience we recall some properties of quasi-greedy expansions in the next section. These properties are also stated in [BK] and are closely related to some important results, first established in the seminal works by Rényi [R] and Parry [P]. Sections 3 and 4 are then devoted to the proof of Theorem 1.3. Theorem 1.4 is proved in Section 5, Theorem 1.5 and Theorem 1.6 are proved in Section 6, and our final Theorems 1.7, 1.8, 1.9 and 1.10 are established in Section Quasi-greedy expansions Let q > 1 be a real number and let m = q 1. In the previous section we defined the quasi-greedy expansion as the largest infinite expansion of x (0, m/(q 1)]. In order to prove that this notion is well defined, we introduce the quasi-greedy algorithm: if a i = a i (x) is already defined for i < n, then a n is the largest element of the set {0,..., m} that satisfies n a i q i < x. Of course, this definition only makes sense if x > 0. In the following proposition we show that this algorithm generates an expansion of x, for all x (0, m/(q 1)]. It follows that the quasi-greedy expansion is generated by the quasi-greedy algorithm. Proposition 2.1. Let x (0, m/(q 1)]. Then, a i x = q i. Proof. If x = m/(q 1), then the quasi-greedy algorithm provides a i = m for all i 1 and the desired equality follows. Suppose that x (0, m/(q 1)). Then, by definition of the quasi-greedy algorithm, there exists an index n such that a n < m. First assume that a n < m for infinitely many n. For any such n, we have by definition n a i 0 < x q i 1 q n. Letting n, we obtain x = a i q i. Next assume there exists a largest n such that a n < m. Then, n a i q + N m i q < x n a i i q + 1 i q, n for each N > n. Hence, i=n+1 i=n+1 m q x n a i i q 1 i q n.

8 UNIQUE EXPANSIONS OF REAL NUMBERS 7 Note that for any q > 1 and 1 q n 1 q n = i=n+1 i=n+1 if and only if q = m + 1. Hence, the existence of a largest n such that a n < m is only possible if q is an integer, in which case a n+i = m for all i 1 and x = n a i q i + i=n+1 m q i, m q i, m q = a i i q. i Now we consider the quasi-greedy expansion (α i ) of x = 1. Note that α 1 = m = q 1. If α n < α 1, then n α i q i + 1 q n 1, by definition of the quasi-greedy algorithm. The following lemma states that this inequality holds for each n 1. Lemma 2.2. For each n 1, the inequality n α i (2.1) q + 1 i q 1 n holds. Proof. The proof is by induction on n. For n = 1, the inequality holds, since α 1 +1 q. Assume the inequality is valid for some n N. If α n+1 < α 1, then (2.1) with n replaced by n + 1 follows from the definition of the quasi-greedy algorithm. If α n+1 = α 1, then the same conclusion follows from the induction hypothesis and the inequality α q. Proposition 2.3. The map q (α i ) is a strictly increasing bijection from the open interval (1, ) onto the set of all infinite sequences satisfying (2.2) α k+1 α k+2... α 1 α 2... for all k 1. Proof. By definition of the quasi-greedy algorithm and Proposition 2.1 the map q (α i ) is strictly increasing. According to the preceding lemma, for every n 1, whence (2.3) n α n+1 q α i q + 1 i q α i n q i + α n+2 q Since (α n+i ) is infinite and (α i ) is the largest infinite sequence satisfying (2.3), inequality (2.2) follows. Conversely, let (α i ) be an infinite sequence satisfying (2.2). Solving the equation α i q i = 1,

9 8 MARTIJN DE VRIES AND VILMOS KOMORNIK we obtain a number q > 1. In order to prove that (α i ) is the quasi-greedy expansion of 1 in base q, it suffices to prove that for each n satisfying α n < α 1, the inequality α i q i 1 q n i=n+1 holds. Starting with k 0 := n and using (2.2), we try to define a sequence satisfying for j = 1, 2,... the conditions k 0 < k 1 < α kj 1+i = α i for i = 1,..., k j k j 1 1 and α kj < α kj k j 1. If we obtain in this way an infinite number of indices, then we have k α i q i j k j 1 α i q kj 1+i 1 q kj i=n+1 < j=1 j=1 ( 1 q kj 1 1 q kj ) = 1 q n. If we only obtain a finite number of indices, then there exists a first nonnegative integer N such that (α kn+i) = (α i ) and we have N k α i q i j k j 1 α i q kj 1+i 1 α i + q kj q kn+i i=n+1 j=1 N j=1 ( 1 q kj 1 1 q kj ) + α i q = 1 kn+i q n. The proofs of the following propositions are almost identical to the proof of Proposition 2.3 and are therefore omitted. Proposition 2.4. Fix q > 1 and denote by (α i ) the quasi-greedy expansion of 1 in base q. Then the map x (a i ) is a strictly increasing bijection from (0, α 1 /(q 1)] onto the set of all infinite sequences, satisfying and 0 a n α 1 for all n 1 (2.4) a n+1 a n+2... α 1 α 2... whenever a n < α 1. Proposition 2.5. Fix q > 1 and denote by (α i ) the quasi-greedy expansion of 1 in base q. Then the map x (b i ) is a strictly increasing bijection from [0, α 1 /(q 1)] onto the set of all sequences, satisfying and 0 b n α 1 for all n 1 (2.5) b n+1 b n+2... < α 1 α 2... whenever b n < α 1. Remarks. A sequence (c i ) is univoque if and only if (c i ) is greedy and (α 1 c i ) is greedy. Hence, Theorem 1.1 is a consequence of Proposition 2.5.

10 UNIQUE EXPANSIONS OF REAL NUMBERS 9 The greedy expansion of x [0, α 1 /(q 1)] is generated by the greedy algorithm: if b i = b i (x) is already defined for i < n, then b n is the largest element of A such that n b i q i x. The proof of this assertion goes along the same lines as the proof of Proposition 2.1. Note that the greedy expansion of a number x (0, α 1 /(q 1)] coincides with the quasi-greedy expansion if and only if the greedy expansion of x is infinite. If the greedy expansion (b i ) of x J \ {0} is finite and b n is its last nonzero element, then the quasi-greedy expansion of x is given by (a i ) = b 1...b n 1 b n α 1 α 2..., as follows from the inequalities (2.2), (2.4) and (2.5). 3. Proof of the necessity part of Theorem 1.3 Let q > 1 be a real number. Lemma 3.1. Let (b i ) = b 1 b 2... be a greedy sequence. Then the truncated sequence b 1...b n 0 is also a greedy sequence, where n 1 is an arbitrary positive integer. Proof. The statement follows at once from Proposition 2.5. Lemma 3.2. Let (b i ) α 1 be a greedy sequence and let N be a positive integer. Then there exists a greedy sequence (c i ) > (b i ) such that c 1...c N = b 1...b N. Proof. Since (b i ) α 1, it follows from (2.5) that b n < α 1 for infinitely many n. Hence, we may assume, by enlarging N if necessary, that b N < α 1. Let j=1 I = {1 i N : b i < α 1 } =: {i 1,..., i n }. Since N I, I. Note that for i r I, N i b ir+j r b ir+j = + 1 q j q j q N ir j=1 because (b i ) is greedy and b ir < α 1. Choose y ir such that b N+i (3.1) < y q i ir α 1 /(q 1) and (3.2) N i r j=1 b ir+j q j + 1 q N ir y i r < 1. b N+i q i < 1 Let y = min {y i1,..., y in } and denote the greedy expansion of y by d 1 d Finally, let (c i ) = b 1...b N d 1 d From (3.1) we infer that (c i ) > (b i ). It remains to show that (c i ) is a greedy sequence, i.e., we need to show that (3.3) c n+i q i < 1 whenever c n < α 1. If c n < α 1 and n N, then (3.3) follows from (3.2). If c n < α 1 and n > N, then (3.3) follows from the fact that (d i ) is a greedy sequence.

11 10 MARTIJN DE VRIES AND VILMOS KOMORNIK Lemma 3.3. Let (b i ) be the greedy expansion of some x [0, α 1 /(q 1)] and suppose that for some n 1, b n > 0 and b n+1 b n+2... > α 1 α Then, (i) There exists a number z > x such that [x, z] U q =. (ii) If b j > 0 for some j > n, then there exists a number y < x such that [y, x] U q =. Proof. (i) Choose a positive integer M > n such that b n+1...b M > α 1...α M n. Applying Lemma 3.2, choose a greedy sequence (c i ) > (b i ) such that c 1...c M = b 1...b M. Then, (c i ) is the greedy expansion of some z > x. If (d i ) is the greedy expansion of some element in [x, z], then (d i ) also begins with b 1...b M and hence In particular, we have that d n > 0 and d n+1...d M > α 1...α M n. d n+1 d n+2... > α 1 α We infer from Theorem 1.1 that [x, z] U q =. (ii) It follows from Lemma 3.1 that (c i ) := b 1...b n 0 is the greedy expansion of some y < x. If (d i ) is the greedy expansion of some element in [y, x], then (c i ) (d i ) (b i ) and d 1...d n = b 1...b n. Therefore, d n+1 d n+2... b n+1 b n+2... > α 1 α 2... and d n = b n > 0. It follows from Theorem 1.1 that [y, x] U q =. Proof of the necessity part of Theorem 1.3. If x U q, then the quasi-greedy expansion (a i ) and the greedy expansion (b i ) of x coincide. Hence, the stronger implication (3.4) a m > 0 = a m+1 a m+2... < α 1 α 2... follows from Theorem 1.1. Suppose now that x U q \U q. According to Theorem 1.1, there exists a smallest positive integer n for which First assume that b n > 0 and b n+1 b n+2... α 1 α b n+1 b n+2... > α 1 α Applying Lemma 3.3 we conclude that b i = 0 for i > n. Hence, the quasi-greedy expansion of x is given by We must show that (a i ) = b 1...b n α 1 α a m > 0 = a m+1 a m+2... α 1 α Instead, we prove the stronger implication (3.4). If m > n, then (3.4) follows from Theorem 1.2 and our assumption q U. If m = n, then (3.4) follows from α 1 = 0 < α 1. Now assume that m < n and a m > 0. Since a m = b m, we have that b m > 0 and by minimality of n, b m+1 b m+2... < α 1 α Equivalently, b m+1...b n α 1 < α 1α 2....

12 UNIQUE EXPANSIONS OF REAL NUMBERS 11 Hence, from which it follows that Moreover, according to Theorem 1.2, proving (3.4). Next assume that b m+1...b n < α 1...α n m, a m+1...a n α 1...α n m. a n+1 a n+2... = α 1 α 2... < α n m+1 α n m+2..., (3.5) b n+1 b n+2... = α 1 α If q is an integer, then (α i ) = α 1 = (q 1) and the implication (3.4) follows from the fact that (a i ) is infinite by definition. If q is a non-integer and (3.5) holds, then (b i ) is infinite and therefore (a i ) = (b i ). Hence, we need to show that the implication (3.6) b m > 0 = b m+1 b m+2... α 1 α 2... holds. If m n, then (3.6) follows from b m+1 b m+2... = α m n+1 α m n+2... α 1 α If m < n, then (3.6) follows from the minimality of n. 4. Proof of the sufficiency part of Theorem 1.3 Fix q U and denote the quasi-greedy expansion of x [0, α 1 /(q 1)] by (a i ) = (a i (x)). Suppose that (4.1) a n+1 a n+2... α 1 α 2... whenever a n > 0. In this section we prove that such an element x belongs to the set U q. It follows from Theorem 1.2 and Proposition 2.3 that the quasi-greedy expansion of 1 in base q satisfies (4.2) α k+1 α k+2... α 1 α 2... for all k 1, and (4.3) α k+1 α k+2... < α 1 α 2... for all k 1. Note that a sequence satisfying (4.2) and (4.3) is automatically infinite. Hence, a sequence satisfying (4.2) and (4.3) is the quasi-greedy expansion of 1 in base q for some q U. The following two lemmas are obtained in [KL3]. Lemma 4.1. If (α i ) is a sequence satisfying (4.2) and (4.3), then there exist arbitrary large integers m such that α m > 0 and (4.4) α k+1...α m < α 1...α m k for all 0 k < m. Lemma 4.2. Let (γ i ) be a sequence satisfying γ k+1 γ k+2... γ 1 γ 2... and γ k+1 γ k+2... γ 1 γ 2... for all k 1, with γ j := γ 1 γ j. If for some n 1, then in fact γ n+1...γ 2n γ 1...γ n (γ i ) = (γ 1...γ n γ 1...γ n ).

13 12 MARTIJN DE VRIES AND VILMOS KOMORNIK Now we are able to prove the sufficiency part of Theorem 1.3. For a fixed base q U, we will distinguish between x J with a finite greedy expansion and x J with an infinite greedy expansion. Lemma 4.3. Fix q U. Suppose that x [0, α 1 /(q 1)] has a finite greedy expansion (b i ) and suppose that the quasi-greedy expansion (a i ) of x satisfies the condition (4.1). Then, x U q. Proof. Note that 0 U q. Hence, we may assume that x (0, α 1 /(q 1)]. If b n is the last nonzero element of (b i ), then (a i ) = b 1...b n α 1 α According to Lemma 4.1, there exists a sequence 1 m 1 < m 2 <, such that (4.4) is satisfied with m = m i for all i 1. We may assume that m i > n for all i 1. Consider for each i 1 the sequence (b i j ), given by Define for i 1, the number x i by (b i j ) = b 1...b n (α 1...α mi α 1...α mi ). x i = j=1 Note that the sequence (x i ) i 1 converges to x as i goes to infinity. It remains to show that x i U q for all i 1. According to Theorem 1.1 it suffices to verify that b i j q j. (4.5) b i m+1b i m+2... < α 1 α 2... whenever b i m < α 1 and (4.6) b i m+1 bi m+2... < α 1α 2... whenever b i m > 0. According to (4.3), α mi+1...α 2mi α 1...α mi. Note that this inequality cannot be an equality, for otherwise it would follow from Lemma 4.2 that (α i ) = (α 1...α mi α 1...α mi ). However, this sequence does not satisfy (4.3) for k = m i. Therefore, Equivalently, α mi+1...α 2mi < α 1...α mi. (4.7) α 1...α mi < α mi+1...α 2mi. If m n, then (4.5) and (4.6) follow from (4.2), (4.4) and (4.7). Now assume that m < n. If b i m < α 1, then b i m+1...b i n = b m+1...b n < b m+1...b n α 1...α n m, where the last inequality follows from the fact that (b i ) is a greedy expansion and b m = b i m < α 1. Hence, b i m+1b i m+2... < α 1 α Suppose that b i m = a m > 0. Since by assumption, a m+1 a m+2... α 1 α 2... and since b i m+1...b i n = a m+1...a n, it suffices to verify that b i n+1 bi n+2... < α n m+1α n m+2....

14 UNIQUE EXPANSIONS OF REAL NUMBERS 13 This is equivalent to (4.8) α n m+1 α n m+2... < (α 1...α mi α 1...α mi ). Since n < m i for each i 1, we infer from (4.4) that and (4.8) follows. α n m+1...α mi < α 1...α mi (n m) Lemma 4.4. Fix q U. Suppose that x [0, α 1 /(q 1)] has an infinite greedy expansion (b i ) and suppose that the quasi-greedy expansion (a i ) of x satisfies the condition (4.1). Then, x U q. Proof. We may assume that x / U q. Note that (a i ) = (b i ), since the greedy expansion of x is infinite by assumption. Since x / U q, there exists a first positive integer n such that b n > 0 and b n+1 b n+2... α 1 α According to (4.1) this last inequality is in fact an equality. As before, let 1 m 1 < m 2 < be a sequence such that (4.4) is satisfied with m = m i for all i 1. Again, we may assume that m i > n for all i 1. Consider for each i 1 the sequence (b i j ), given by and define for i 1, the number y i by (b i j ) = b 1...b n (α 1...α mi α 1...α mi ) y i = j=1 Then the sequence (y i ) i 1 converges to x as i goes to infinity. It remains to show that y i U q for all i 1. If m n, then (4.5) and (4.6) follow as in the proof of the preceding lemma. Now assume that m < n. If b i m = b m < α 1, then by (2.5), Hence, it suffices to verify that b i j q j. b i m+1...b i n = b m+1...b n α 1...α n m. b i n+1 bi n+2... = (α 1...α mi α 1...α mi ) < α n m+1 α n m+2..., which is already done (cf. (4.8)). Finally, suppose that b i m = b m > 0. We must verify that b i m+1...bi nb i n+1... < α 1α By minimality of n, we have i.e., b m+1...b n b n+1... < α 1 α 2..., (4.9) b m+1...b n α 1 α 2... < α 1 α Therefore, b i m+1...bi n = b m+1...b n < α 1...α n m, for otherwise b m+1...b n = α 1...α n m and it would follow from (4.9) that which contradicts (4.2). α 1 α 2... < α n m+1 α n m+2...,

15 14 MARTIJN DE VRIES AND VILMOS KOMORNIK 5. Proof of Theorem 1.4 In order to prove Theorem 1.4 we start with some preliminary lemmas. Lemma 5.1. Fix q > 1. If (b i ) α 1 is the greedy expansion of a number x J, i.e., if 0 x < α 1 /(q 1), then there exists a sequence 1 n 1 < n 2 < such that for each i 1, (5.1) b ni < α 1 and b m+1...b ni < α 1...α ni m if m < n i and b m < α 1. Proof. We define a sequence (n i ) i 1 satisfying the requirements by induction. Let r be the first positive integer for which b r < α 1. Then, (5.1) with n i replaced by r holds trivially. Set n 1 := r. Suppose we have already defined n 1 < < n l, such that for each 1 j l, b nj < α 1 and b m+1...b nj < α 1...α nj m if m < n j and b m < α 1. Since (b i ) is greedy and b nl < α 1, there exists a first integer n l+1 > n l such that (5.2) b nl+1...b nl+1 < α 1...α nl+1 n l. Note that b nl+1 < α nl+1 n l α 1. It remains to verify that for all 1 m < n l+1 for which b m < α 1, we have that (5.3) b m+1...b nl+1 < α 1...α nl+1 m. If m < n l, then (5.3) follows from the induction hypothesis. If m = n l, then (5.3) reduces to (5.2). If n l < m < n l+1, then b nl+1...b m = α 1...α m nl, by minimality of n l+1, and thus by (5.2), b m+1...b nl+1 < α m nl+1...α nl+1 n l α 1...α nl+1 m. The following lemma has been established in [KL3]: Lemma 5.2. If q U \ U, then the greedy expansion (β i ) of 1 is finite and all expansions of 1 are given by (5.4) (α i ) and (α 1...α m ) N α 1...α m 1 α + m 0, N = 0, 1, 2,..., where m is such that β m is the last nonzero element of (β i ). Note that if the greedy expansion (β i ) of 1 is finite with last nonzero element β m, then the quasi-greedy expansion of 1 is given by (α i ) = (β 1...β m) = (α 1...α m ). Hence, as a consequence of Lemma 5.2, for each q U, the quasigreedy expansion (α i ) of 1 is also the smallest expansion of 1 in lexicographical order. Proof of Theorem 1.4. (ia) We establish that U q \ U q = ℵ 0. More specifically, if q U \ N, then the sets A q and B q (introduced in a remark following the statement of Theorem 1.4) are countably infinite. Moreover, the greedy expansion of a number x B q ends with α 1 α If q {2, 3,...}, then A q = U q \ U q. Fix q U. Denote the greedy expansion of a number x U q \ U q by (b i ). Since x / U q, there exists a number n such that b n > 0 and b n+1 b n+2... α 1 α If this inequality is strict, then b i = 0 for all i > n (cf. Lemma 3.3). Otherwise, the sequence (b i ) ends with α 1 α 2..., which is infinite unless q is an integer. It follows from Theorem 1.1, Theorem 1.2 and Theorem 1.3 that a sequence of the form 0 n 10 for n 0, is the finite greedy expansion of 1/q n+1 U q \U q. Moreover,

16 UNIQUE EXPANSIONS OF REAL NUMBERS 15 if q U \ N, then a sequence of the form α n 1 α 1α 2... for n 1, is the infinite greedy expansion of a number x U q \ U q. These observations conclude the proof. (ib) We show that if q U, then A q is dense in U q. Fix q U. For each x U q, we will define a sequence (x i ) i 1 of numbers in A q U q \ U q that converges to x. We have seen in the proof of part (i) that 1/q n A q for each n 1. Hence, there exists a sequence of numbers in A q that converges to 0 U q. Suppose now that x U q \ {0} and denote by (c i ) the unique expansion of x. Since c 1 c 2... α 1 is greedy, we infer from Lemma 5.1 that there exists a sequence 1 n 1 < n 2 <, such that for each i 1, (5.5) c ni > 0 and c m+1...c ni < α 1...α ni m if m < n i and c m > 0. Now consider for each i 1 the sequence (b i j), given by and define the number x i by (b i j) = c 1...c ni 0 x i = j=1 According to Lemma 3.1, the sequences (b i j ) are the finite greedy expansions of the numbers x i, i 1. Moreover, the sequence (x i ) i 1 converges to x as i goes to infinity. We claim that x i A q for each i 1. Note that x i / U q because the quasi-greedy expansion (a i j ), given by b i j q j. c 1...c n i α 1 α 2... is another expansion of x i. According to Theorem 1.3, it remains to prove that (5.6) a i j > 0 = a i j+1 ai j+2... α 1α If j < n i and a i j > 0, then a i j+1...ai n i = c j+1...c n i α 1...α ni j, by (5.5) and a i n i+1 ai n... = α i+2 1α 2... < α ni j+1α ni j+2..., by Theorem 1.2. If j = n i, then (5.6) follows from α 1 = 0 < α 1. Finally, if j > n i, then (5.6) follows again from Theorem 1.2. (ic) We show that if q U \ N, then the set B q is dense in U q. Fix q U \ N. For each x U q, we will define a sequence (x i ) i 1 of numbers in B q U q \ U q that converges to x. It follows from Theorem 1.1, Theorem 1.2 and Theorem 1.3 that a sequence of the form 0 i α 1 α 1 α 2... for i 0, is the infinite greedy expansion of a number x i U q \ U q. Note that the sequence (x i ) i 1 converges to 0 U q. Therefore, we may again assume that x U q \ {0}. Let (c i ) be the unique expansion of a number x U q \ {0}, and let 1 n 1 < n 2 < be a sequence of integers satisfying (5.5). Arguing as in the proof of part (ib), one finds that for each i 1, the sequence (b i j) = c 1...c ni α 1 α 2... is the infinite greedy expansion of a number x i U q \ U q. (ii) and (iii) Fix q U and let (b i ) be the greedy expansion of some number x U q \ U q. Let n be the smallest positive integer for which b n > 0 and b n+1 b n+2... α 1 α Let (d i ) be another expansion of x. Then, (d i ) < (b i ) and hence there exists a smallest integer j for which d j < b j. First we show that j n. Suppose by contradiction that j < n. Then, b j > 0 and by minimality of n, we have b j+1 b j+2... > α 1 α 2....

17 16 MARTIJN DE VRIES AND VILMOS KOMORNIK Since α 1 α 2... is the smallest expansion of 1, we have that α 1 α 2... is the largest expansion of the number i 1 α i/q i, and therefore, But then, d j+i q i = b j+i q i > > = b j+i q i α i q i + 1 α 1 q i, α i q i. + b j d j which is clearly impossible. If j = n, then d n = b n, for otherwise we have again d n+i b n+i q i q i + 2 > b n+i q i + α 1 q i, α i q + α i i q i where the second inequality follows from the fact that (α i ) is the smallest expansion of 1 and the inequality α 1 α 2... < α 1 α Now we distinguish between two cases: If j = n and (5.7) b n+1 b n+2... > α 1 α 2..., then by Lemma 3.3, we have b r = 0 for r > n, from which it follows that (d n+i ) is an expansion of 1. Hence, if q U and (5.7) holds, then the only expansion of x starting with b 1...b n is given by (c i ) := b 1...b n α 1 α If q U \ U and (5.7) holds, then any expansion (c i ) starting with b 1...b n is an expansion of x if and only if (c n+i ) is one of the expansions listed in (5.4). If j = n and (5.8) b n+1 b n+2... = α 1 α 2..., then d n+i q i = b n+i q i + 1 = Hence, if (5.8) holds, then the only expansion of x starting with b 1...b n is given by b 1...b n α 1. Finally, if j > n, then b n+1 b n+2... = α 1 α 2..., for otherwise (b n+i ) = 0 and d j < b j is impossible. Note that in this case q / U, because otherwise (b n+i ) is the unique expansion of i 1 α i/q i and thus (d n+i ) = (b n+i ) which is impossible since j > n. Hence, if q U, then (b i ) is the only expansion of x starting with b 1...b n. If q U \ U and (5.8) holds, then any α 1 q i.

18 UNIQUE EXPANSIONS OF REAL NUMBERS 17 expansion (c i ) starting with b 1...b n is an expansion of x if and only if (c n+i ) is one of the conjugates of the expansions listed in (5.4). The statements of parts (ii) and (iii) follow directly from the above considerations. Remark. Fix q U. It follows from Theorem 1.4 (i) that each x U q \U q has either a finite expansion or an expansion that ends with α 1 α 2..., i.e., x can be written as x = b 1 q + + b n q + 1 ( ) α1 n q n q 1 1. Moreover, according to Lemma 5.2, the greedy expansion of 1 in base q is finite if q U\U. Hence, if q U is transcendental, then each x U q \U q is a transcendental number. If q U is an algebraic number or if q U \ U, then each x U q \ U q is also algebraic. 6. Proof of Theorem 1.5 and Theorem 1.6 Fix q > 1. It follows from Proposition 2.3 and Proposition 2.5 that a sequence (b i ) is greedy if and only if 0 b n α 1 for all n 1, and (6.1) b n+k+1 b n+k+2... < α 1 α 2... for all k 0, whenever b n < α 1. Lemma 6.1. Assume that q / U. Then a greedy sequence (b i ) cannot end with α 1 α Proof. Assume by contradiction that for some n, b n+1 b n+2... = α 1 α Since in this case b n+1 = α 1 = 0 < α 1, it would follow from (6.1) that α k+1 α k+2... < α 1 α 2... for all k 1. But this contradicts our assumption that q / U. Lemma 6.2. Assume that q / U. Then, (i) The set U q is closed. (ii) Each element x V q \ U q has a finite greedy expansion. Proof. (i) Let x J \ U q and denote the greedy expansion of x in base q by (b i ). According to Theorem 1.1, there exists a positive integer n such that b n > 0 and b n+1 b n+2... α 1 α Applying Lemma 3.3 and Lemma 6.1 we conclude that [x, z] U q =, for some number z > x. It follows that U q is closed from above. Note that the set U q is symmetric in the sense that x U q α 1 /(q 1) x U q, as follows from Theorem 1.1. Hence, the set U q is also closed from below. (ii) Let x V q \ U q and suppose that (a i (x)) = (b i (x)). Then, it would follow that for some positive integer n, contradicting Lemma 6.1. b n+1 b n+2... = α 1 α 2..., Lemma 6.3. Let (a i ) be the quasi-greedy expansion of some x [0, α 1 /(q 1)]. Furthermore, let M be an arbitrary positive integer. Then a 1...a M 0 is a greedy sequence.

19 18 MARTIJN DE VRIES AND VILMOS KOMORNIK Proof. If x = 0, then there is nothing to prove. If x 0, then the statement follows from Proposition 2.4 and Proposition 2.5. Recall from the introduction that the set V consists of those numbers q > 1 for which the quasi-greedy expansion (α i ) of 1 in base q satisfies (6.2) α k+1 α k+2... α 1 α 2... for all k 1. Note that the quasi-greedy expansion of 1 in base q for q V \ U is of the form (6.3) (α i ) = (α 1...α k α 1...α k ), where k 1 is the first integer for which equality holds in (6.2). In particular, such a sequence is periodic. Any sequence of the form (1 n 0 n ), where n is a positive integer, is infinite and satisfies (2.2) and (6.2) but not (4.3). On the other hand, there are only countably many periodic sequences. Hence, the set V \U is countably infinite. Note that α k > 0, for otherwise it would follow from (6.2) and (6.3) that α k α k+1...α 2k 1 = α 1 α 1 α 2...α k 1 α 1...α k 1 0, which is impossible, because α 1 > 0. The following lemma ([KL3]) implies that the number of expansions of 1 is countably infinite in case q V \ U. Moreover, all expansions of the number 1 in base q are determined explicitly. Lemma 6.4. If q V \ U, then all expansions of 1 are given by (α i ), and the sequences (α 1...α 2k ) N α 1...α 2k 1 α + 2k 0, N = 0, 1,... and (α 1...α 2k ) N α 1...α k 1 α k α 1, N = 0, 1,.... It follows from the above lemma that for each q V \ U, the greedy expansion of 1 in base q is given by (β i ) = α 1...α 2k 1 α + 2k 0 and the smallest expansion of 1 in base q is given by α 1...α k 1 α k α 1. Now we are ready to prove Theorem 1.5 and Theorem 1.6. Throughout the proof of Theorem 1.5, q V \ U is fixed but arbitrary, and k 1 is the first positive integer for which equality holds in (6.2). Proof of Theorem 1.5. (i) We show that the sets U q and V q are closed. In view of Lemma 6.2, it remains to prove that V q is closed. Fix x J \ V q and let (a i (x)) = (a i ) be the quasi-greedy expansion of x. Then, there exists an integer n > 0, such that Let m be such that a n > 0 and a n+1 a n+2... > α 1 α (6.4) a n+1...a n+m > α 1...α m, and let y = n+m According to Lemma 6.3, the greedy expansion of y is given by a 1...a n+m 0. Therefore, the quasi-greedy expansion of each number v (y, x] starts with the block a 1...a n+m. It follows from (6.4) that Consider now the sequence a i q i. (y, x] V q =. (d i ) = a 1...a n α 1 α 2....

20 UNIQUE EXPANSIONS OF REAL NUMBERS 19 It follows from (2.4) and (6.2) that (d i ) is the quasi-greedy expansion of d i /q i = z > x. i 1 Note that the quasi-greedy expansion (v i ) of an element v [x, z) satisfies Hence, from which the claim follows. v n = a n > 0 and v n+1 v n+2... < α 1 α [x, z) V q =, (iia) We prove that V q \ U q = ℵ 0. The set V q \ U q is countable, because each element x V q \ U q has a finite greedy expansion (cf. Lemma 6.2). On the other hand, for each n 1, the sequence α n 1 0 is the greedy expansion of an element x V q \ U q, from which the claim follows. (iib) In order to show that V q \ U q is dense in V q, one can argue in the same way as in the proof of Theorem 1.4 (ib). Instead of applying Theorem 1.2 one should now apply the inequality (6.2). (iic) Finally, we show that all elements of V q \ U q are isolated points of V q. Let x V q \ U q and let b n be the last nonzero element of the greedy expansion (b i ) of x. Choose m such that α m < α 1. Note that this is possible because q / N. According to Lemma 3.2, there exists a greedy sequence (c i ) > (b i ), such that If we set c 1...c n+m = b 1...b n 0 m. z = then the quasi-greedy expansion (v i ) of a number v (x, z] starts with b 1...b n 0 m. Hence, v n = b n > 0, and Therefore, c i q i, v n+1...v n+m = α m 1 > α 1...α m. (x, z] V q =. In order to show that there exists also a number y < x, such that (y, x) V q =, we introduce for m 1 the sequences (b m j ), given by and we define the numbers x m by (b m j ) = b 1...b n (α 1...α k α 1...α k ) m 0 x m = j=1 b m j q j. The sequences (b m j ) are all greedy by Lemma 6.3. Moreover, x m x as m goes to infinity. Let v (x m, x m+1 ] for some m 1, and let the quasi-greedy expansion of v be given by (d i ). Then, and Therefore, d 1...d n d n+1...d 2km+n = b 1...b n(α 1...α k α 1...α k ) m d 2km+n+1...d 2k(m+1)+n < α 1...α k α 1...α k. d 2k(m 1)+n+k = α k > 0

21 20 MARTIJN DE VRIES AND VILMOS KOMORNIK and Hence, v / V q, i.e., d 2k(m 1)+n+k+1...d 2k(m+1)+n > α 1...α k α 1...α k α 1...α k = α 1...α 3k. (x 1, x) V q =. (iii) We already know from Lemma 6.2 that each x V q \ U q has a finite greedy expansion. It remains to show that each element x V q \ U q has ℵ 0 expansions. Let x V q \ U q and let b n be the last nonzero element of its greedy expansion (b i ). If j < n and b j = a j > 0, then because x V q. Therefore, a j+1...a n = b j+1...b n α 1...α n j, (6.5) b j+1...b n > α 1...α n j. Let (d i ) be another expansion of x and let j be the smallest positive integer for which d j b j. Since (b i ) is greedy, we have d j < b j and j {1,..., n}. First we show that j {n k, n}. Suppose by contradiction that j / {n k, n}. First assume that n k < j < n. Then, b j > 0 and by (6.5), b j+1...b n 0 > α 1...α n j α n j+1...α k 0. Since α 1...α k α 1 is the smallest expansion of 1 in base q, α 1...α k 0 is the greedy expansion of α 1 /(q 1) 1, and therefore, b j+i q i > α 1 /(q 1) 1. But then, d j+i q i = b j+i q i > α 1 /(q 1), + b j d j which is impossible. Next assume that 1 j < n k. Rewriting (6.5), one gets If then Hence, b j+1...b n < α 1...α n j. b j+1...b j+k = α 1...α k, b j+k+1...b n < α k+1...α n j. b j+k+1 b j+k+2... > α k+1 α k+2... = α 1 α Since in this case b j+k = α k < α 1, the last inequality contradicts the fact that (b i ) is a greedy sequence. Hence, if j < n k, then Equivalently, b j+1...b j+k < α 1...α k. b j+1...b j+k α 1...α k.

22 UNIQUE EXPANSIONS OF REAL NUMBERS 21 Since n > j + k and b n > 0, it follows that b j+1 b j+2... > α 1...α k 0, which leads to the same contradiction as at the beginning of the proof. It remains to investigate what happens if j {n k, n}. If j = n k, then it follows from (6.5) that Equivalently, and thus (6.6) b n k+1...b n α 1...α k. b n k+1 b n k+2... = b n k+1...b n 0 α 1...α k 0, d n k+i q i b n k+i q i + 1 α 1 /(q 1), where both inequalities in (6.6) are equalities if and only if d n k = b n k, b n k+1...b n = α 1...α k, and d n k+1d n k+2... = α 1. Hence, d n k < b n k is only possible in case b n k > 0 and b n k+1...b n = α 1...α k. Finally, if j = n, then d n = b n, for otherwise d n+i α i q i 2 > q i + α i q i = α 1/(q 1), because α 1 α 2... < α 1...α k α 1. In this case (d n+i ) is one of the expansions listed in Lemma 6.4. Remark. Fix q V \ U. According to Lemma 6.4, the number 1 has a finite greedy expansion in base q. Hence, each element q V \ U is algebraic. Because each x V q \ U q has a finite greedy expansion in base q, it follows that the set V q \ U q consists entirely of algebraic numbers. Lemma 6.5. Let (α i ) be the quasi-greedy expansion of 1 in some base q > 1 and assume there exists a positive integer k such that α k+1 α k+2... > α 1 α 2... Then there exists a positive integer m such that α m > 0 and α m+1 α m+2... > α 1 α Proof. Let m = max {1 i k : α m > 0}. Note that m is well defined, since α 1 > 0. Then, α m+1...α k = Hence, α m+1 α m+2... > α 1 α Proof of Theorem 1.6. Fix q / V. In view of Lemma 6.2, it remains to prove that a number x J \ {0} with a finite greedy expansion does not belong to V q. Let x J \ {0} be an element with a finite greedy expansion. Since q / V, there exists a positive integer k, such that α k+1 α k+2... > α 1 α According to Lemma 6.5, we may assume that α k > 0. Since the quasi-greedy expansion of each element x J \ {0} with a finite greedy expansion ends with α 1 α 2..., we conclude that x / V q.

23 22 MARTIJN DE VRIES AND VILMOS KOMORNIK 7. Proof of Theorem 1.7 Theorem 1.10 In this section we will complete our study of the univoque set U q for numbers q > 1. The results proved in the preceding sections were mainly concerned with various properties of the sets U q for numbers q V. Now we will use these properties to describe the topological structure of U q for all numbers q > 1. Since the set V is closed, we may write (1, )\V as the union of countably many disjoint open intervals (q 1, q 2 ): the connected components of (1, ) \ V. In order to determine the endpoints of these components we recall from [KL3] that V is closed. V \ U is dense in V. all elements of V \ U are isolated in V. In fact, for each element q U there exists a sequence (q m ) m 1 of numbers in V \ U such that q m q, as can be seen from the proof of Theorem 2.6 in [KL3]. Proposition 7.1. (i) The set R of right endpoints q 2 of the connected components (q 1, q 2 ) is given by R = V \ U. (ii) The set L of left endpoints q 1 of the connected components (q 1, q 2 ) is given by L = N (V \ U). Proof of Proposition 7.1 (i). Note that V \ U R because the set V \ U is discrete. As we have already observed in the preceding paragraph, each element q U can be approximated from below by elements in V \ U. Hence, R = V \ U. The proof of part (ii) of Proposition 7.1 requires more work. We will prove a number of technical lemmas first. In the remainder of this section q (α i ) indicates that the quasi-greedy expansion of 1 in base q is given by (α i ). For convenience we also write 1 0, and we occasionally refer to 0 as the quasi-greedy expansion of the number 1 in base 1. Let q 2 V \ U and suppose that where k is chosen to be minimal. q 2 (α i ) = (α 1...α k α 1...α k ) Remark. The minimality of k implies that the smallest period of (α i ) equals 2k. Indeed, if j is the smallest period of (α i ), then α j = α 2k = α k < α 1 because j divides 2k. Hence, α 1...α + j 0 is an expansion of 1 in base q 2 which contradicts Lemma 6.4 if j < 2k. Lemma 7.2. For all 0 i < k, we have α i+1...α k < α 1...α k i. Proof. For i = 0, the inequality follows from the relation α 1 = 0 < α 1. Hence, assume that 1 i < k. Since q 2 V, Suppose that for some 1 i < k, If k 2i, then α i+1...α k α 1...α k i. α i+1...α k = α 1...α k i. α 1...α 2i = α 1...α i α 1...α i, and it would follow from Lemma 4.2 that (α i ) = (α 1...α i α 1...α i ),

24 UNIQUE EXPANSIONS OF REAL NUMBERS 23 contradicting the minimality of k. If i < k < 2i, then α i+1...α 2i = α i+1...α k α 1...α 2i k = α 1...α k i α 1...α 2i k leading to the same contradiction. α 1...α k i α k i+1...α i = α 1...α i, Let q 1 be the largest element of V {1} that is smaller than q 2. This element exists because the set V {1} is closed and the elements of V \ U are isolated points of V {1}. The next lemma provides the quasi-greedy expansion of 1 in base q 1. Lemma 7.3. q 1 (α 1...α k ). Proof. Let q 1 (v i ). If k = 1, then q 2 (α 1 0) and (v i ) = (α 1 ) because q 2 is the smallest element of V (α 1, α 1 + 1). Hence, we may assume that k 2. Observe that v 1...v k α 1...α k. If v 1...v k = α 1...α k, then i.e., v k+1...v 2k α 1...α k, v k+1...v 2k α 1...α k, = v 1...v k (here we need that k 2, for otherwise the conjugate bars on both sides have a different meaning) and it would follow from Lemma 4.2 that q 1 = q 2. Hence, v 1...v k α 1...α k. It follows from Proposition 2.3 that (w i ) = (α 1...α k ) is the largest quasi-greedy expansion of 1 in some base q > 1 that starts with α 1...α k. Therefore, it suffices to show that the sequence (w i ) satisfies inequality (6.2) for all k 1. Since the sequence (w i ) is periodic with period k, it suffices to verify that (7.1) w j+1 w j+2... w 1 w 2... for all 0 j < k. If j = 0, then (7.1) is true because w 1 = 0 < w 1 ; hence assume that 1 j < k. Then, according to the preceding lemma, and Hence, so that α j+1...α k < α 1...α k j, α 1...α j < α k j+1...α k. w j+1...w j+k = α j+1...α k α 1...α j α 1...α k j α 1...α j < α 1...α k, w j+1...w j+k w 1...w k. Since the sequence (w j+i ) = w j+1 w j+2... is also periodic with period k, the inequality (7.1) follows.

25 24 MARTIJN DE VRIES AND VILMOS KOMORNIK Lemma 7.4. Fix q > 1 and denote by (β i ) the greedy expansion of the number 1 in base q. For any positive integer n, we have Proof. It follows from (6.1) that β n+1 β n+2... β 1 β β n+1 β n+2... < α 1 α 2... β 1 β 2..., whenever there exists a positive integer j n satisfying β j < β 1 = α 1. If such an integer j does not exist, then either (β i ) = α 1 or there exists an integer j > n for which β j < α 1. In both these cases the desired inequality readily follows as well. Now we consider a number q 1 V \ U. Recall from Lemma 5.2 and Lemma 6.4 that the greedy expansion (β i ) of 1 in base q 1 is finite. Denote its last nonzero element by β m. Lemma 7.5. (i) The smallest element q 2 of V that is larger than q 1 exists. Moreover, q 2 (β 1...β m β 1...β m ). (ii) The greedy expansion of 1 in base q 2 is given by (γ i ) = β 1...β m β 1...β m0. Proof. (i) First of all, note that q 1 (α i ) = (β 1...β m). Moreover, (β 1...βm ) is the largest quasi-greedy expansion of 1 in some base q > 1 that starts with β 1...βm. Hence, in view of Lemma 4.2, it suffices to show that the sequence satisfies the inequalities (w i ) = (β 1...β m β 1...β m ) (7.2) w k+1 w k+2... w 1 w 2... and (7.3) w k+1 w k+2... w 1 w 2... for all k 0. Observe that (7.2) for k + m is equivalent to (7.3) for k and (7.3) for k + m is equivalent to (7.2) for k. Since both relations are obvious for k = 0, we only need to verify (7.2) and (7.3) for 1 k < m. Fix 1 k < m. The relation (7.3) follows from our assumption that q 1 V: w k+1...w m = β k+1...β m < α k+1...α m α 1...α m k = w 1...w m k. Since 1 m k < m, we also have Using Lemma 7.4, we obtain from which (7.2) follows. (ii) We must show that w m k+1...w m < w 1...w k. w k+1...w k+m = w k+1...w m w 1...w k w 1...w m k w 1...w k < w 1...w m k w m k+1...w m, (7.4) γ k+1 γ k+2... < w 1 w 2... whenever γ k < w 1.